Question 562516
as the clock strikes noon, Jack is 250 yards from the starting point and Edson is 400 yards from the start.
 Jack jogs at the constant pace of 3 yards/sec. 
Edson jogs at the constant pace of 2.5 yards/sec.
 How long will it take Jack to catch up to Edson?
:
From the information given, Ed has 400-250 = 150 yd head start (at noon)
:
Let t = time required for Jack to catch Ed
:
Write a distance equation, dist = speed * time
:
J's dist = Ed's dist + 150 yds
3t = 2.5t + 150
3t - 2.5t = 150
.5t = 150
Multiply both sides by 2
t = 300 seconds or 5 min, 12:05 would be the time
:
:
Check this by finding the dist each ran 
3(300) = 900 yds
2.5(300)=750 yds
----------------
differs: 150 yds